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				<span style="position: absolute;left:15px;bottom:15px;width:90%;"><font class="view-text" style="color:#fcfcfc;font-size:25px">题解 P5400 【[CTS2019]随机立方体】</font><br><a href="/tags/2021/" class="tag"><span  style="background-color: rgb(52, 152, 219);">2021</span></a>&nbsp;<a href="/tags/二项式反演/" class="tag"><span  style="background-color: rgb(231, 76, 60);">二项式反演</span></a>&nbsp;<a href="/tags/题解/" class="tag"><span  style="background-color: rgb(82, 196, 26);">题解</span></a></span>
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                <h2 id="_1">题意</h2>
<p>有一个 <script type="math/tex">n\times m\times l</script> 的立方体，立方体中每个格子上都有一个数，如果某个格子上的数是所在的三个平面内最大的，我们就称它是极大的。</p>
<p>现在将 <script type="math/tex">1\sim n\times m\times l</script> 这 <script type="math/tex">n\times m\times l</script> 个数等概率随机填入 <script type="math/tex">n\times m\times l</script> 个格子，使得每个数恰出现一次，求恰有 <script type="math/tex">k</script> 个极大的数的概率。答案对 <script type="math/tex">998244353</script> 取模。</p>
<!--more-->
<h2 id="_2">题解</h2>
<p>一个 <del>非常有趣</del> 的<strong>神仙题</strong>。</p>
<p>
<script type="math/tex">3</script> 维的题目就是毒瘤</p>
<p>首先恰好并不好做，考虑<strong>钦定</strong>+二项式反演。</p>
<p>记 <script type="math/tex">dp_i</script> 表示<strong>钦定</strong> <script type="math/tex">i</script> 个极大点的方案数，有：
<script type="math/tex; mode=display">dp_i=\sum_{j=i}\binom{j}{i}ans_j\Leftrightarrow ans_i=\sum_{j=i}(-1)^{j-i}\binom{j}{i}dp_j</script>
这个 <script type="math/tex">dp</script> 看上去也不是很好求，那么一步步来分析。</p>
<p>首先若是<strong>钦定</strong>的极大点数量已知，那么于这些点共面的点也就自然已知，记为 <script type="math/tex">g_i</script>。显然一个极大点会占用一行、一列、一层，没有两个极大点共面。因此剩下的是 <script type="math/tex">(n-i)(m-i)(l-i)</script>，即：
<script type="math/tex; mode=display">g_i=nml-(n-i)(m-i)(l-i)</script>
设总数为 <script type="math/tex">N=nml</script>，那么我们<strong>钦定</strong>  <script type="math/tex">i</script> 个极大值会有 <script type="math/tex">g_i</script> 个位置有限制，选出这些数的方案为 <script type="math/tex">\binom{N}{g_i}</script>。剩下 <script type="math/tex">N-g_i</script> 个位置放任自流，为 <script type="math/tex">(N-g_i)!</script>。</p>
<p>但是现在我们无法区分两个选择数相同、摆放不同的方案。我们记 <script type="math/tex">f_i</script> 为 <script type="math/tex">i</script> 个极大值的摆放方案，因为坐标互不影响，只需两两互不相同，由乘法原理得到：
<script type="math/tex; mode=display">f_i=n^{\underline{i}}m^{\underline{i}}l^{\underline{i}}</script>
于是我们现在只剩下摆放那 <script type="math/tex">g_i</script> 个数的方案。此时不用管是那些数，也不用管在那些位置。因为交换行、列、层不影响公共部分（即约束条件）。</p>
<p>我们完全可以认为我们要求的是这么摆放的：</p>
<p><img alt="" src="/img/eflhqjxh.png"></p>
<p>并且有 <script type="math/tex">\color{green}\text{绿}\color{black}\text{块}>\color{blue}\text{蓝}\color{black}\text{块}>\color{yellow}\text{黄}\color{black}\text{块}>\color{red}\text{红}\text{块}</script>
</p>
<p>考虑只有红块，那么方案数为 <script type="math/tex">(g_1-1)!</script>，因为最大值必定在红块。记为这个值 <script type="math/tex">h_1</script>。</p>
<p>此时若是加入黄块，显然必定可以加入且成为极大点。会新增 <script type="math/tex">g_2-g_1</script> 个位置。这些位置怎么放不会影响红块。此时方案数为 <script type="math/tex">(g2-1)^{\underline{g_2-g_1-1}}\times h_1</script>。把最大值放在黄块上，选 <script type="math/tex">g_2-g_1-1</script> 个值放在多出来的位置。剩下 <script type="math/tex">g_1</script> 个就沿用 <script type="math/tex">h_1</script> 的方案，即 <script type="math/tex">h_2=\frac{(g_2-1)!}{g_1!}h_1</script>。</p>
<p>用类似的方法可以推出 <script type="math/tex">h_3=\frac{(g_3-1)!}{g_2!}h_2</script>。更一般的，有：
<script type="math/tex; mode=display">\begin{aligned}
h_i&=\frac{(g_i-1)!}{g_{i-1}!}h_{i-1}\\
&=\prod_{j=1}^i\frac{(g_j-1)!}{g_{j-1}!}\\
&=\frac{\prod_{j=1}^ig_j!}{\prod_{j=1}^ig_j\prod_{j=1}^ig_{j-1}!}\\
&=\frac{\sout{\prod_{j=1}^{i-1}g_j!}\times g_i!}{\prod_{j=1}^ig_j\sout{\prod_{j=1}^{i-1}g_j!}}\\
&=g_i!\prod_{j=1}^i\frac{1}{g_j}
\end{aligned}</script>
</p>
<p>不难发现 <script type="math/tex">h_i</script> 就是 <script type="math/tex">i</script> 个极大值时的方案数了。</p>
<p>于是就可以写出 <script type="math/tex">dp_i</script> 的表达式了：
<script type="math/tex; mode=display">
\begin{aligned}
dp_i&=\binom{N}{g_i}(N-g_i)!f_ih_i\\
&=\frac{N!}{g_i!(N-g_i)!}(N-g_i)!f_ig_i!\prod_{j=1}^i\frac1{g_j}\\
&=N!f_i\prod_{j=1}^i\frac1{g_j}
\end{aligned}
</script>
由于最后求的是概率，我们完全可以在这一步就把 <script type="math/tex">N!</script> 除掉，得到 <script type="math/tex">\hat{dp_i}=f_i\prod_{j=1}^i\frac1{g_j}</script>
</p>
<p>于是写出最终的柿子：
<script type="math/tex; mode=display">\frac{\sum_{j=k}(-1)^{j-k}\binom{j}{k}dp_j}{N!}</script>
<script type="math/tex; mode=display">\sum_{j=k}(-1)^{j-k}\binom{j}{k}\hat{dp_j}</script>
最后就是几个实现的问题了：</p>
<ul>
<li>
<p>计算 <script type="math/tex">f</script> 时可以递推</p>
</li>
<li>
<p>
<script type="math/tex">\prod_{j=1}^i\frac1{g_j}</script> 需要用线性求法</p>
</li>
</ul>
<p>实现起来并不是很难。</p>
<h2 id="_3">代码</h2>
<p>这里仅给出核心代码。
<div class="highlight"><pre><span></span><code><span class="linenos" data-linenos=" 1 "></span><span class="kt">int</span> <span class="n">t</span><span class="p">;</span><span class="n">mint</span> <span class="n">fac</span><span class="p">[</span><span class="n">N</span><span class="p">],</span><span class="n">ifac</span><span class="p">[</span><span class="n">N</span><span class="p">];</span>
<span class="linenos" data-linenos=" 2 "></span><span class="n">mint</span> <span class="n">g</span><span class="p">[</span><span class="n">N</span><span class="p">],</span><span class="n">f</span><span class="p">[</span><span class="n">N</span><span class="p">],</span><span class="n">dp</span><span class="p">[</span><span class="n">N</span><span class="p">];</span>
<span class="linenos" data-linenos=" 3 "></span><span class="n">mint</span> <span class="n">prod</span><span class="p">[</span><span class="n">N</span><span class="p">],</span><span class="n">inv</span><span class="p">[</span><span class="n">N</span><span class="p">];</span>
<span class="linenos" data-linenos=" 4 "></span><span class="n">mint</span> <span class="nf">C</span><span class="p">(</span><span class="kt">int</span> <span class="n">n</span><span class="p">,</span><span class="kt">int</span> <span class="n">m</span><span class="p">){</span><span class="k">return</span> <span class="n">fac</span><span class="p">[</span><span class="n">n</span><span class="p">]</span><span class="o">*</span><span class="n">ifac</span><span class="p">[</span><span class="n">m</span><span class="p">]</span><span class="o">*</span><span class="n">ifac</span><span class="p">[</span><span class="n">n</span><span class="o">-</span><span class="n">m</span><span class="p">];}</span>
<span class="linenos" data-linenos=" 5 "></span><span class="kt">void</span> <span class="nf">work</span><span class="p">(){</span>
<span class="linenos" data-linenos=" 6 "></span>    <span class="kt">int</span> <span class="n">n</span><span class="p">,</span><span class="n">m</span><span class="p">,</span><span class="n">l</span><span class="p">,</span><span class="n">k</span><span class="p">;</span>
<span class="linenos" data-linenos=" 7 "></span>    <span class="n">in</span><span class="o">::</span><span class="n">read</span><span class="p">(</span><span class="n">n</span><span class="p">,</span><span class="n">m</span><span class="p">,</span><span class="n">l</span><span class="p">,</span><span class="n">k</span><span class="p">);</span>
<span class="linenos" data-linenos=" 8 "></span>    <span class="n">mint</span> <span class="n">N</span><span class="o">=</span><span class="n">mint</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span><span class="o">*</span><span class="n">n</span><span class="o">*</span><span class="n">m</span><span class="o">*</span><span class="n">l</span><span class="p">;</span>
<span class="linenos" data-linenos=" 9 "></span>    <span class="kt">int</span> <span class="n">mi</span><span class="o">=</span><span class="n">min</span><span class="p">(</span><span class="n">n</span><span class="p">,</span><span class="n">min</span><span class="p">(</span><span class="n">m</span><span class="p">,</span><span class="n">l</span><span class="p">));</span>
<span class="linenos" data-linenos="10 "></span>    <span class="n">f</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span>
<span class="linenos" data-linenos="11 "></span>    <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span><span class="n">i</span><span class="o">&lt;=</span><span class="n">mi</span><span class="p">;</span><span class="n">i</span><span class="o">++</span><span class="p">)</span>
<span class="linenos" data-linenos="12 "></span>        <span class="n">f</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">=</span><span class="n">f</span><span class="p">[</span><span class="n">i</span><span class="mi">-1</span><span class="p">]</span><span class="o">*</span><span class="p">(</span><span class="n">n</span><span class="o">-</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">m</span><span class="o">-</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">l</span><span class="o">-</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">);</span>
<span class="linenos" data-linenos="13 "></span>    <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="n">i</span><span class="o">&lt;=</span><span class="n">mi</span><span class="p">;</span><span class="n">i</span><span class="o">++</span><span class="p">)</span>
<span class="linenos" data-linenos="14 "></span>        <span class="n">g</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">=</span><span class="n">N</span><span class="o">-</span><span class="n">mint</span><span class="p">(</span><span class="mi">1</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">n</span><span class="o">-</span><span class="n">i</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">m</span><span class="o">-</span><span class="n">i</span><span class="p">)</span><span class="o">*</span><span class="p">(</span><span class="n">l</span><span class="o">-</span><span class="n">i</span><span class="p">);</span>
<span class="linenos" data-linenos="15 "></span>    <span class="n">prod</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span><span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span><span class="n">i</span><span class="o">&lt;=</span><span class="n">mi</span><span class="p">;</span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="n">prod</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">=</span><span class="n">prod</span><span class="p">[</span><span class="n">i</span><span class="mi">-1</span><span class="p">]</span><span class="o">*</span><span class="n">g</span><span class="p">[</span><span class="n">i</span><span class="p">];</span>
<span class="linenos" data-linenos="16 "></span>    <span class="n">inv</span><span class="p">[</span><span class="n">mi</span><span class="p">]</span><span class="o">=</span><span class="mi">1</span><span class="o">/</span><span class="n">prod</span><span class="p">[</span><span class="n">mi</span><span class="p">];</span><span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="n">mi</span><span class="p">;</span><span class="n">i</span><span class="p">;</span><span class="n">i</span><span class="o">--</span><span class="p">)</span><span class="n">inv</span><span class="p">[</span><span class="n">i</span><span class="mi">-1</span><span class="p">]</span><span class="o">=</span><span class="n">inv</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">*</span><span class="n">g</span><span class="p">[</span><span class="n">i</span><span class="p">];</span>
<span class="linenos" data-linenos="17 "></span>    <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span><span class="n">i</span><span class="o">&lt;=</span><span class="n">mi</span><span class="p">;</span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="n">dp</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">=</span><span class="n">f</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">*</span><span class="n">inv</span><span class="p">[</span><span class="n">i</span><span class="p">];</span>
<span class="linenos" data-linenos="18 "></span>    <span class="n">mint</span> <span class="n">ans</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span>
<span class="linenos" data-linenos="19 "></span>    <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">j</span><span class="o">=</span><span class="n">k</span><span class="p">;</span><span class="n">j</span><span class="o">&lt;=</span><span class="n">mi</span><span class="p">;</span><span class="n">j</span><span class="o">++</span><span class="p">)</span>
<span class="linenos" data-linenos="20 "></span>        <span class="k">if</span><span class="p">((</span><span class="n">j</span><span class="o">-</span><span class="n">k</span><span class="p">)</span><span class="o">&amp;</span><span class="mi">1</span><span class="p">)</span><span class="n">ans</span><span class="o">-=</span><span class="n">C</span><span class="p">(</span><span class="n">j</span><span class="p">,</span><span class="n">k</span><span class="p">)</span><span class="o">*</span><span class="n">dp</span><span class="p">[</span><span class="n">j</span><span class="p">];</span>
<span class="linenos" data-linenos="21 "></span>        <span class="k">else</span> <span class="n">ans</span><span class="o">+=</span><span class="n">C</span><span class="p">(</span><span class="n">j</span><span class="p">,</span><span class="n">k</span><span class="p">)</span><span class="o">*</span><span class="n">dp</span><span class="p">[</span><span class="n">j</span><span class="p">];</span>
<span class="linenos" data-linenos="22 "></span>    <span class="n">out</span><span class="o">::</span><span class="n">write</span><span class="p">(</span><span class="n">ans</span><span class="p">.</span><span class="n">x</span><span class="p">);</span>
<span class="linenos" data-linenos="23 "></span>    <span class="n">out</span><span class="o">::</span><span class="n">putc</span><span class="p">(</span><span class="sc">&#39;\n&#39;</span><span class="p">);</span> 
<span class="linenos" data-linenos="24 "></span><span class="p">}</span>
<span class="linenos" data-linenos="25 "></span><span class="kt">signed</span> <span class="nf">main</span><span class="p">(){</span>
<span class="linenos" data-linenos="26 "></span>    <span class="n">fac</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span><span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span><span class="n">i</span><span class="o">&lt;</span><span class="n">N</span><span class="p">;</span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="n">fac</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">=</span><span class="n">fac</span><span class="p">[</span><span class="n">i</span><span class="mi">-1</span><span class="p">]</span><span class="o">*</span><span class="n">i</span><span class="p">;</span>
<span class="linenos" data-linenos="27 "></span>    <span class="n">ifac</span><span class="p">[</span><span class="n">N</span><span class="mi">-1</span><span class="p">]</span><span class="o">=</span><span class="mi">1</span><span class="o">/</span><span class="n">fac</span><span class="p">[</span><span class="n">N</span><span class="mi">-1</span><span class="p">];</span><span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="n">N</span><span class="mi">-1</span><span class="p">;</span><span class="n">i</span><span class="p">;</span><span class="n">i</span><span class="o">--</span><span class="p">)</span><span class="n">ifac</span><span class="p">[</span><span class="n">i</span><span class="mi">-1</span><span class="p">]</span><span class="o">=</span><span class="n">ifac</span><span class="p">[</span><span class="n">i</span><span class="p">]</span><span class="o">*</span><span class="n">i</span><span class="p">;</span>
<span class="linenos" data-linenos="28 "></span>    <span class="n">in</span><span class="o">::</span><span class="n">read</span><span class="p">(</span><span class="n">t</span><span class="p">);</span><span class="k">while</span><span class="p">(</span><span class="n">t</span><span class="o">--</span><span class="p">)</span><span class="n">work</span><span class="p">();</span>
<span class="linenos" data-linenos="29 "></span>    <span class="n">out</span><span class="o">::</span><span class="n">flush</span><span class="p">();</span>
<span class="linenos" data-linenos="30 "></span>    <span class="k">return</span> <span class="mi">0</span><span class="p">;</span>
<span class="linenos" data-linenos="31 "></span><span class="p">}</span>
</code></pre></div></p>
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